Generate All Permutations of Array in DSA Go - Build from Scratch

Generate All Permutations of Array

📖 Scenario: You are working on a puzzle game where you need to find all possible ways to arrange a set of unique numbers. This helps the game generate different challenges for players.

🎯 Goal: Build a Go program that generates all permutations of a given array of integers and prints each permutation.

📋 What You'll Learn

Create an array called nums with the values 1, 2, 3

Create a variable called result to store all permutations as a slice of slices

Write a recursive function called permute that generates all permutations of nums

Print each permutation in result as a slice

💡 Why This Matters

🌍 Real World

Generating permutations is useful in games, puzzles, and testing all possible configurations of a set of items.

💼 Career

Understanding recursion and permutations is important for algorithm design, coding interviews, and solving complex problems in software development.

1

Create the initial array

Create an array called nums with the exact values 1, 2, 3.

DSA Go

# Your code here

Hint

Use nums := []int{1, 2, 3} to create the array.

2

Create a variable to store permutations

Create a variable called result as a slice of slices of integers to store all permutations. Initialize it as empty.

DSA Go

package main

func main() {
    nums := []int{1, 2, 3}
    // Your code here
}

Hint

Use var result [][]int to declare an empty slice of slices.

3

Write the recursive permutation function

Write a recursive function called permute that takes nums []int, start int, and a pointer to result *[][]int. It should generate all permutations of nums by swapping elements starting from start. When start reaches the length of nums, append a copy of nums to *result. Call permute(nums, 0, &result) inside main.

DSA Go

package main

func main() {
    nums := []int{1, 2, 3}
    var result [][]int
    // Your code here
}

// Write permute function below

Hint

Use swapping and recursion to generate permutations. Remember to backtrack by swapping back.

4

Print all permutations

Add a for loop in main to print each permutation in result as a slice using fmt.Println.

DSA Go

package main

import "fmt"

func permute(nums []int, start int, result *[][]int) {
    if start == len(nums) {
        temp := make([]int, len(nums))
        copy(temp, nums)
        *result = append(*result, temp)
        return
    }
    for i := start; i < len(nums); i++ {
        nums[start], nums[i] = nums[i], nums[start]
        permute(nums, start+1, result)
        nums[start], nums[i] = nums[i], nums[start]
    }
}

func main() {
    nums := []int{1, 2, 3}
    var result [][]int
    permute(nums, 0, &result)
    // Your code here
}

Hint

Use for _, perm := range result { fmt.Println(perm) } to print each permutation.

Practice

1. What does generating all permutations of an array mean?

easy

A. Finding the largest element in the array

B. Sorting the array in ascending order

C. Listing all possible orders of the array elements

D. Removing duplicate elements from the array

Solution

1. Step 1: Understand permutations

   Permutations are all possible ways to arrange elements in order.

2. Step 2: Compare options

   Only Listing all possible orders of the array elements describes listing all possible orders of elements.

3. Final Answer:

   Listing all possible orders of the array elements -> Option C

4. Quick Check:

   Permutations = all orders [OK]

Hint: Permutations = all ways to arrange elements [OK]

Common Mistakes:

• Confusing permutations with sorting
• Thinking permutations remove duplicates
• Mixing permutations with finding max element

2. Which Go code snippet correctly starts a recursive function to generate permutations of an integer slice arr?

easy

A. func permute(arr int[], start int) {}

B. func permute(arr []int, start int) {}

C. func permute(arr []int, start string) {}

D. func permute(arr []int) {}

Solution

1. Step 1: Check Go function syntax

   Go uses []int for slices and int for integers; parameters must match expected types.

2. Step 2: Validate parameters for permutation

   We need the array slice and an integer index to track recursion start.

3. Final Answer:

   func permute(arr []int, start int) {} -> Option B

4. Quick Check:

   Correct Go slice and int params [OK]

Hint: Use []int for slices and int for indices [OK]

Common Mistakes:

• Using incorrect slice syntax like int[]
• Using wrong parameter types like string for index
• Omitting necessary parameters for recursion

3. What will be the output of this Go code snippet that generates permutations of []int{1, 2} and prints them?

func permute(arr []int, start int) {
  if start == len(arr) {
    fmt.Println(arr)
    return
  }
  for i := start; i < len(arr); i++ {
    arr[start], arr[i] = arr[i], arr[start]
    permute(arr, start+1)
    arr[start], arr[i] = arr[i], arr[start]
  }
}

func main() {
  permute([]int{1, 2}, 0)
}

medium

A. Compilation error

B. [2 1]
[1 2]

C. [1 2]

D. [1 2]
[2 1]

Solution

1. Step 1: Trace recursion for arr = [1, 2]

   Start=0: swap arr[0] with arr[0] -> [1,2], recurse start=1; then swap arr[0] with arr[1] -> [2,1], recurse start=1.

2. Step 2: Print when start == len(arr)

   At start=2, print current arr. So prints [1 2] then [2 1].

3. Final Answer:

   [1 2] [2 1] -> Option D

4. Quick Check:

   All permutations printed in order [OK]

Hint: Swap and backtrack to print all permutations [OK]

Common Mistakes:

• Not swapping back after recursion
• Printing before full permutation formed
• Confusing order of printed permutations

4. Identify the bug in this Go function intended to generate all permutations of arr:

func permute(arr []int, start int) {
  if start == len(arr) {
    fmt.Println(arr)
    return
  }
  for i := 0; i < len(arr); i++ {
    arr[start], arr[i] = arr[i], arr[start]
    permute(arr, start+1)
    arr[start], arr[i] = arr[i], arr[start]
  }
}

medium

A. Loop should start from 'start' instead of 0

B. Missing base case for recursion

C. Swapping elements is incorrect

D. Function should return a slice, not print

Solution

1. Step 1: Analyze loop range

   The loop runs from i=0 to len(arr), but it should start from 'start' to avoid duplicate swaps.

2. Step 2: Understand impact of loop start

   Starting from 0 causes repeated permutations and incorrect swapping order.

3. Final Answer:

   Loop should start from 'start' instead of 0 -> Option A

4. Quick Check:

   Loop start index = start [OK]

Hint: Loop from start index to avoid duplicates [OK]

Common Mistakes:

• Looping from zero causing repeated permutations
• Not swapping back after recursion
• Ignoring base case correctness

5. Given an array []int{1, 2, 3}, how many unique permutations will be generated by a correct permutation function, and why?

hard

A. 6, because 3! (3 factorial) = 6 permutations

B. 3, because there are 3 elements

C. 9, because 3 * 3 = 9 combinations

D. 1, because the array is fixed

Solution

1. Step 1: Calculate permutations count for 3 elements

   Number of permutations = factorial of number of elements = 3! = 3*2*1 = 6.

2. Step 2: Understand why other options are incorrect

   3 is just count of elements, 9 is product not permutations, 1 is incorrect as array can reorder.

3. Final Answer:

   6, because 3! (3 factorial) = 6 permutations -> Option A

4. Quick Check:

   Permutations of 3 elements = 6 [OK]

Hint: Permutations count = factorial of array length [OK]

Common Mistakes:

• Confusing permutations count with array length
• Using multiplication instead of factorial
• Thinking only one arrangement exists